Root Locus from equation?

19 次查看（过去 30 天）
Ali Almakhmari
Ali Almakhmari 2023-11-4
回答： Paul 2024-11-15，3:39
I have this equation:
syms s u
eqn = 1472.*s.^4 - 99.2.*s.^3 -256.*s.^2.*u.^2 + 1393.*s.^2 + 2.4.*s.*u.^2 - 25.*s -24.*u.^2 + 150 == 0;
where u is speed, and s are the eigenvalues of the systems. I want to plot the root locus (real vs imaginary part of the eigenvalues) as a function of the speed but I am not sure where to start because the root locus command in MATLAB doesnt take something like this.

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2023-11-4
Ran in:
It might if you aske it to and give it a causal system it can work with —
s = tf('s');
eqn = @(u) 1 / (1472*s^4 - 99.2*s^3 -256*s^2*u.^2 + 1393*s^2 + 2.4*s*u^2 - 25*s -24*u^2 + 150)
eqn = function_handle with value:
@(u)1/(1472*s^4-99.2*s^3-256*s^2*u.^2+1393*s^2+2.4*s*u^2-25*s-24*u^2+150)
u = 1;
figure
rlocus(eqn(u))
grid
u = 10;
figure
rlocus(eqn(u))
grid
The original ‘eqn’ would only work for fimplicit, for example —
eqn = @(s,u) 1472.*s.^4 - 99.2.*s.^3 -256.*s.^2.*u.^2 + 1393.*s.^2 + 2.4.*s.*u.^2 - 25.*s -24.*u.^2 + 150;
figure
fimplicit(eqn, [[-1 1]*10 [-1 1]*10])
grid
I would not consider that to be a root locus plot, however that is likely the only way to work with it as originally stated.
.

更多回答（1 个）

Paul
Paul 2024-11-15，3:39
Ran in:
In order to use rlocus, we need to get the characteristic equation in a form of 1 + u^2*N(s)/D(s) = 0.
Start with the equation
syms s u
eqn = 1472.*s.^4 - 99.2.*s.^3 -256.*s.^2.*u.^2 + 1393.*s.^2 + 2.4.*s.*u.^2 - 25.*s -24.*u.^2 + 150 == 0;
Sub in u2 = u^2
syms u2
eqn2 = subs(eqn,u^2,u2)
Find the coefficients of u2 and 1
[C,T] = coeffs(lhs(eqn2),u2,'All')
Now eqn can be expressed as 1 + u2*N(s)/D(s) = 0, where
N(s) = C(1);
D(s) = C(2);
Convert to numeric
N = sym2poly(N(s));
D = sym2poly(D(s));
and plot the root locus. Here, the root locus "gain" is u^2
figure
rlocus(tf(N,D))
We can verify by first having rlocus return the roots of the eqn for each value of u^2
[r,u2] = rlocus(tf(N,D));
and then verify that eqn is satisfied for the first (or second/third/fourth) root returned for the values of u^2.
figure
plot(abs(double(subs(lhs(eqn),[s,u^2],{r(1,:).',u2.'}))))

类别

在 Help Center 和 File Exchange 中查找有关 Classical Control Design 的更多信息

标签

产品


版本

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by